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The game show host puts seven cards with the numbers 1, 2, 4, 8, 16, 32, 64 into a bowler hat. The candidate picks cards up randomly from the hat, one card at a time, and without putting cards back, until the sum of the numbers on cards in his hand exceeds 124.

What is the most probable sum the candidate can get in the end? On which sum should he bet?

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The following sets of cards exceed $124$:

  • $64 + 32 + 16 + 8 + 4 + 2 + 1 = 127$
  • $64 + 32 + 16 + 8 + 4 + 2 = 126$
  • $64 + 32 + 16 + 8 + 4 + 1 = 125$

Thus, you must always pick at least $6$ cards to exceed $124$. The unpicked card in this case must be a $1$ or a $2$, otherwise you will need to pick all $7$ cards.

This means that there is a $\frac{2}{7}$ chance that you will have a sum less than $127$. The most likely value, therefore, is $127$.

The expected value $E$ is:

$$\begin{align}E & = 127 \times \frac{5}{7} + 126 \times \frac{1}{7} + 125 \times \frac{1}{7} \\ & = \frac{127 \times 5 + 126 + 125}{7} \\ & = \frac{886}{7} \\ & = 126 \frac{4}{7} \end{align}$$

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    $\begingroup$ Note that this problem is equivalent to the following: the candidate takes all seven cards, shuffles them, and the flips them over one at a time until the sum exceeds 124. The only way for the final score to not be 127 is if the cards 1 or 2 are on the bottom of the deck. (Visualizing it this way helped me understand your "unpicked card" logic.) $\endgroup$ – Michael Seifert Mar 10 '16 at 15:10
  • $\begingroup$ @MichaelSeifert Great analogy! $\endgroup$ – Trenin Mar 10 '16 at 15:24
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The player has to pick at least $6$ cards whatever number on the card is at the beginning since the sum of biggest numbers does not exceed $124$ until $6$ cards are picked.

If the player picks $4$ or $8$ or $16$ or $32$ or $64$ as a the last card, he has to pick all cards since the total would not exceed $124$ until the last card ispicked!
The chance of the last card being any of these numbers is $\frac{5}{7}$ and the sum would be $127$ for these cases.

So $\frac{5}{7} > 50 \%$, you do not have to calculate the rest and the most probably sum would be

$127$

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  • $\begingroup$ excellent way to think of it $\endgroup$ – Kevin Mar 10 '16 at 15:38
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You should bet on

127

because

it is five times better than the either of the other choices. All the matters is what the last card is. If the last card is the 1, you will not draw it and your sum will be 126. If your last card is the 2, you will not draw it and your sum will be 125. If your last card is any other card, you will draw all 7 and your sum will be 127. So, you have a $\frac{1}{7}$ chance of getting 125, a $\frac{1}{7}$ chance of getting 126 and a $\frac{5}{7}$ chance of getting 127.

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